f(Ø) = 21Ø + 21(sinØ)^2
f ' (Ø) = 21 + 42sinØcosØ = 21 + 21sin (2Ø)
f '' (Ø) = 21 cos(2Ø) (2)
= 42 cos (2Ø)
the period of cos 2Ø is π , and our domain is [0,π]
so we have one complete cosine curve to consider.
Remember that a function is concave up if its second derivative is positive, and negative if .....
42 cos(2Ø) is positive from 0 to π/4 and then again from 3π/4 to π
so it is concave up for 0 < Ø < π/4 OR 3π/4 < Ø < π
and concave down between π/4 and 3π/4
Determine the intervals on which the function is concave up and concave down
f(theta) = 21theta + 21Sin^2(theta),
[0,pi]
I have tried to implicitly differentiate but I am getting nowhere,... can anyone please help?
2 answers
why implicit?
f(x) = 21x + 21sin^2 x
f' = 21 + 21 sin 2x
f'' = 42 cos 2x
f''=0 at x=pi/4,3pi/4
f''>0 for x in (0,pi/4)
f''<0 for x in (pi/4,3pi/4)
f''>0 for x in (3pi/4,pi)
so now you know the cioncavity.
f(x) = 21x + 21sin^2 x
f' = 21 + 21 sin 2x
f'' = 42 cos 2x
f''=0 at x=pi/4,3pi/4
f''>0 for x in (0,pi/4)
f''<0 for x in (pi/4,3pi/4)
f''>0 for x in (3pi/4,pi)
so now you know the cioncavity.
3 answers